Getting the derivatives of the inverse trigonometric functions

0:00This topic opens with a promise that is withdrawn in the same breath. These functions are continuous, it says, and no proof will be offered. Which is a strange thing to skip, because what happens next is stronger. We are about to differentiate them, and a function that has a derivative somewhere is automatically continuous there. So rather than take continuity on trust, this video measured it. At ninety-nine inputs spread across the stretch, the angle you approach is the angle you arrive at.

0:37And at both ends of that stretch as well. Hold on to those two ends. By the time we finish, they will be the only two inputs in the whole topic that lose their derivative. Start with the inverse sine. Call the input x, and call the angle it returns y. So y is the angle whose sine is x. There is no formula for that angle, and there is not going to be one.

1:05But there is something better: read the sentence backwards. If y is the angle whose sine is x, then x is the sine of y. That is no longer an inverse function. It is a relation between two letters, and a relation can be differentiated exactly as it stands. Nothing has been solved. The sentence has only been turned round. Differentiate both sides with respect to the input. The left side is the input itself, so its derivative is one.

1:41The right side is the sine of something that is itself a function of the input, so it needs the chain rule. One factor per stage: the cosine of the angle, times the derivative of the angle. And the derivative of the angle is exactly what we are hunting. So: one equals the cosine of the angle, times the thing we want. Divide, and the hunt is over in one line.

2:08The derivative of the inverse sine is one over the cosine of the angle. Stop on that answer for a moment, because it is easy to walk past. It is a real answer. It is also an untidy one. The inverse function is still sitting inside it, buried in a cosine. It always gets improved, and we will improve it too. But notice what this form has before we lose it.

2:37It contains no choice at all. A cosine has one value at one angle. There is nothing to decide. Remember that, because in three minutes it will be the whole point. First, though, ask where this answer fails. It is a fraction, so it fails where the bottom is nought. The cosine of the angle is nought only at the two ends of the stretch the inverse sine lives on.

3:06Those two ends are the angles whose sines are minus one and one. So exactly two inputs are lost: minus one, and one. And here is the part worth saying out loud. The function is perfectly well defined at those two inputs. The angle whose sine is one exists. It is the quarter turn, and this video found it by searching rather than by quoting it. It is only the derivative that fails there, and it fails hard.

3:37Squeeze the neighbourhood ten times narrower and the difference quotient climbs again, past ten thousand. Every whole answer from minus ten to ten was offered to it and refused. So the stretch the derivative lives on is smaller than the stretch the function lives on, by exactly two inputs. At all ninety-nine inputs in between, the derivative answers every time. Now let us improve the answer and get rid of that buried inverse.

4:11The sine of the angle is the input. That is where we started. Square it, and use the identity that ties a sine to a cosine. The square of the cosine of the angle is one, less the square of the input. That step is completely safe, and it was checked to be safe. On all four stretches where the sine is one-to-one, at every input tried, the square of the cosine really is one less the square.

4:41Thirty-six checks, thirty-six agreements. So far, nothing has been chosen. And now the one decision in the whole topic. We know the square of the cosine. We want the cosine. A square root has two signs, and the mathematics does not tell you which. The usual sentence is: the angle lies in the stretch from minus a quarter turn to a quarter turn, the cosine is positive there, so take the positive root.

5:13That sentence is correct, and it is doing far more work than it looks. It is not a formality. It is the only thing standing between you and a wrong sign. Here is how to see that. The sine is one-to-one on the stretch from minus a quarter turn to a quarter turn, so an inverse sine exists there. But the sine is also one-to-one on the next stretch up, from a quarter turn to three quarters of a turn.

5:43And on the one after that. And on the one below. Four stretches, four inverse sines, all of them honest. So put the same nine inputs to all four, and measure the derivative of each from its own difference quotient. One over the root of one less the square is the answer for two of the four stretches, nine inputs each. For the other two it is wrong at every single input.

6:12And the other sign of the root answers for exactly the two it missed. Which two, and why? Look at where the cosine sits on each stretch. Positive, negative, positive, negative. The winning sign is the cosine's sign, every time. Now go back to the untidy answer, the one with the angle still inside it. One over the cosine of the angle is the derivative on all four stretches, nine inputs each, thirty-six for thirty-six.

6:46So the sign problem was not created by the mathematics. It was created by the rewriting. A cosine has one sign. A root has two. The choice appears at the root and nowhere before it. With that settled, here is the table you actually need. The inverse sine: one over the root of one less the square, for inputs strictly between minus one and one. The inverse cosine: the same thing with a minus in front, on the same inputs.

7:21The inverse tangent: one over one plus the square, for every input there is. Three rows. Copy them if you like, but the next two minutes are better spent watching two of them get derived, because the derivation is three lines and it never leaves you. A student who can produce the second row will never mis-sign it, and mis-signing it is the commonest mistake in this whole area. The inverse cosine, then, with the same three moves.

7:55The angle whose cosine is the input. Read it backwards: the input is the cosine of the angle. Differentiate: one equals minus the sine of the angle, times the derivative. The minus arrives from the derivative of the cosine, and it is going to survive to the end. Isolate: minus one over the sine of the angle. Square, rewrite, and the root appears again, with the same two signs. This time the stretch runs from nought to a half turn, and on that stretch the sine is positive.

8:32So the root is positive again, and the minus in front is what you keep. Measured on three stretches where the cosine is one-to-one: the negative answer wins on that one, the positive answer on the other two, nine inputs each. Same argument, different ratio, opposite sign. The inverse tangent looks like more of the same, and it is not. Backwards: the input is the tangent of the angle. Differentiate: one equals the square of the secant of the angle, times the derivative.

9:10Now the identity: the square of the secant is one plus the square of the tangent. And the tangent of the angle is the input. So the square of the secant is one plus the square of the input. The derivative is one over one plus the square. Read that again and notice what did not happen. No root was taken. No sign was chosen. And the measurement says the same: that answer is right on all three stretches where the tangent is one-to-one, nine inputs each, and its negative on none of them.

9:49Where the inverse sine's answer had to be told which stretch it was on, this one does not. That is also why its inputs are every number there is, with nothing lost at either end. Out to five hundred either way, it answers. Three rows, and there are six inverse trigonometric functions. So three of them get no derivative here at all: the inverse cotangent, the inverse secant, the inverse cosecant.

10:20That is a fact about the table, not about the mathematics. All three have derivatives, and the same three moves find them. The inverse cotangent: minus one over one plus the square. The inverse secant: one over the input times the root of the square less one. The inverse cosecant: the same with a minus. Each was measured on its own stretches, and each had its opposite sign offered and refused.

10:50So if you meet one of these, you are not stuck. You are one derivation away. There is an easier route still, and it is worth knowing because it turns up in the exercises. The secant is one over the cosine. So the angle whose secant is the input is the angle whose cosine is one over the input. Measured at ten inputs: the same angle, every time. Likewise the angle whose cosecant is the input is the angle whose sine is one over the input.

11:27And the angle whose cotangent is the input is a quarter turn less the angle whose tangent is the input. So an expression written with a ratio the table leaves out can always be rewritten with one it carries. Nothing you will be asked actually needs a row that is not there. But a student who misses the substitution has nothing to fall back on, which is a real cost.

11:55Which brings us to the shape of almost every exercise in this area. You are handed an inverse function of something ugly. A fraction, or a root, or both. The instinct is to differentiate it where it stands, and that is a long and painful road. The move is to substitute first. Let the input be the tangent of an angle, or the cosine of one, and watch the ugly bracket collapse.

12:25Twice the input over one plus the square becomes the sine of twice the angle. One less the square over one plus the square becomes the cosine of twice the angle. So the whole expression turns into a plain multiple of a single inverse function, and then the table finishes it in one step. Seven such items were worked here. Two collapse to twice the inverse tangent, one to three times it, two to a quarter turn less twice it, one to twice the inverse sine, and one to twice the inverse cosine.

13:04At all twenty-two inputs the collapse was checked as an equality of angles, and then the answer was checked against the item's own difference quotient. Twenty-two for twenty-two, with the same answers half again as large refused every time. Every one of those items comes with an interval attached, and it is the thing most likely to be ignored. It is not the domain. It is the condition under which the collapse is true at all.

13:37So the test was run the other way as well: step outside the interval and see what happens. In six of the seven items, every outside input tried gives a different angle. The collapse is simply false out there. The seventh is more interesting, and it is worth being exact about. It fails for negative inputs, as expected. But above the interval it keeps working, on inputs the stated condition excludes.

14:08So for that one item the stated interval is narrower than it needs to be. The lesson survives either way: check the interval, because six times out of seven leaving it makes your answer wrong. Two items deserve a slower look. The first is the inverse sine of the input times its own square root. That inner expression is the input to the power three halves, so the chain rule gives three halves times the root of the input, over the root of one less the cube.

14:44Right at three inputs inside, with the same answer half again as large refused at each. And at the far end, where that lower root vanishes, nothing at all is accepted. The second is written with an inverse cotangent of a ratio of two square roots, which looks impossible with a table that has no cotangent row. Substitute, and it collapses to half the input. Its derivative is a half, at all three inputs tried, and a third was offered and refused.

15:19A ratio the table never carries, finished with what the table does carry. Here is the single best item in this area. Add two angles: the angle whose sine is the input, and the angle whose sine is the root of one less the square. Differentiate the sum. Term by term is perfectly legal. The first term gives the positive root answer, the second gives the negative one, and they cancel exactly.

15:52The derivative is nought. Both terms were measured separately, and their two measured bands add to nought at all five inputs tried. But look again, because there is a one-line route. The second angle is just the angle whose cosine is the input. Measured, at all five: the same angle. And an angle whose sine is the input, plus an angle whose cosine is the input, is a quarter turn.

16:21A constant. Constants have derivative nought, and you are done before you started. One last item, to show where these derivatives are going. Take an exponential whose exponent is a constant times the angle whose cosine is the input. Differentiate once, using the row we derived: you get minus the constant, times the function itself, over the root of one less the square. That was measured at fourteen pairs of constant and input, with the opposite sign refused at all fourteen and the same answer half again as large refused too.

17:05Multiply through by that root to clear the denominator, square both sides, differentiate again, and divide by twice the first derivative. What drops out is a relation between the function, its first derivative and its second. One less the square, times the second derivative, less the input times the first, less the square of the constant times the function, is nought. And that relation was scored against three rivals: one with a sign flipped in the middle, one with a sign flipped at the end, one with the leading factor dropped.

17:46Every second derivative in that test was measured, never quoted. The relation we derived can be nought at all fourteen pairs. The three rivals, between them, at two. And those two are both the case where the input is nought. Which is the old warning in a new place: test a relation at nought and three of the four candidates pass. Test it anywhere else and only one survives. So, the whole topic in five moves.

18:19Read the inverse backwards, so the inverse sign disappears and a relation appears. Differentiate both sides, and isolate the derivative you want. Read the lost inputs off the bottom of the fraction, and say them out loud, because the function keeps them and the derivative does not. Rewrite the ratio of the angle as a root, and then stop and choose the sign, using the stretch the inverse was defined on.

18:50And when the expression is ugly, substitute before you differentiate. Three of the six inverse ratios come with a derivative here, three do not, and the three that do not are one substitution away from the three that do. The one thing to carry out of all this is the sign step. Everything else in this topic is bookkeeping. That one line is an argument, and it is the only place where the choice made when the inverse was first defined does any visible work.